Answer:
2. cotФ = [tex]-\frac{4}{3}[/tex]
3. sec²Ф - tan²Ф = 1
4. [tex]\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}[/tex] = cos²∝
5. The value of sin x is [tex]\frac{1}{2}[/tex]
6. cos 15° = [tex]\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]
7. tan(α + β) = [tex]-\frac{63}{16}[/tex]
8. sin(π - Ф) = sin Ф
9. cos 2Ф = [tex]-\frac{1}{2}[/tex]
10. sin 2Ф = 0.96
Step-by-step explanation:
2.
∵ cos Ф = [tex]-\frac{4}{5}[/tex]
∵ 90° < Ф < 180°
- That means Ф lies on the 2nd quadrant
∴ sin Ф is a positive value
∵ sin²Ф + cos²Ф = 1
∴ sin²Ф + ( [tex]-\frac{4}{5}[/tex] )² = 1
∴ sin²Ф + [tex]\frac{16}{25}[/tex] = 1
- Subtract from both sides [tex]\frac{16}{25}[/tex]
∴ sin²Ф = [tex]\frac{9}{25}[/tex]
- Take √ for both sides
∴ sinФ = [tex]\frac{3}{5}[/tex]
∵ cotФ = cosФ ÷ sinФ
∴ cotФ = [tex]-\frac{4}{5}[/tex] ÷ [tex]\frac{3}{5}[/tex]
∴ cotФ = [tex]-\frac{4}{3}[/tex]
3.
The expression is sec²Ф - tan²Ф
∵ tan²Ф = sec²Ф - 1
- Substitute tan²Ф by the right hand side in the expression
∴ sec²Ф - tan²Ф = sec²Ф - (sec²Ф - 1)
∴ sec²Ф - tan²Ф = sec²Ф - sec²Ф + 1
- Simplify the right hand side
∴ sec²Ф - tan²Ф = 1
4.
The expression is [tex]\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}[/tex]
∵ The numerator is sin²∝ + cos²∝
∵ sin²∝ + cos²∝ = 1
∴ The numerator = 1
∵ The denominator is tan²∝ + 1
∵ tan²∝ = [tex]\frac{sin^{2}\alpha}{cos^{2}\alpha}[/tex]
∴ tan²∝ + 1 = [tex]\frac{sin^{2}\alpha}{cos^{2}\alpha}[/tex] + 1
- Change 1 to fraction [tex]\frac{cos^{2}\alpha}{cos^{2}\alpha}[/tex]
∴ tan²∝ + 1 = [tex]\frac{sin^{2}\alpha}{cos^{2}\alpha}[/tex] + [tex]\frac{cos^{2}\alpha}{cos^{2}\alpha}[/tex]
- Add the two fractions
∴ tan²∝ + 1 = [tex]\frac{sin^{2}\alpha+cos^{2}\alpha }{cos^{2}\alpha}[/tex]
∵ sin²∝ + cos²∝ = 1
∴ tan²∝ + 1 = [tex]\frac{1}{cos^{2}\alpha}[/tex]
∴ The denominator = [tex]\frac{1}{cos^{2}\alpha}[/tex]
∴ [tex]\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}[/tex] = [tex]\frac{1}{\frac{1}{cos^{2}\alpha}}[/tex]
- Remember denominator the denominator will be a numerator
∴ [tex]\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}[/tex] = cos²∝
5.
∵ tan x cos x = [tex]\frac{1}{2}[/tex]
∵ tan x = [tex]\frac{sin(x)}{cos(x)}[/tex]
∴ [tex]\frac{sin(x)}{cos(x)}[/tex] × cos x = [tex]\frac{1}{2}[/tex]
- Simplify it by canceling cos x up with cos x down
∴ sin x = [tex]\frac{1}{2}[/tex]
∴ The value of sin x is [tex]\frac{1}{2}[/tex]
6.
∵ cos(Ф - ∝) = cosФ cos∝ + sinФ sin∝
∵ 45 - 30 = 15
∴ cos 15° = cos(45 - 30)°
- Use the rule above to find the exact value
∵ cos(45 - 30)° = cos 45° cos 30° + sin 45° sin 30°
∵ cos 45° = [tex]\frac{\sqrt{2}}{2}[/tex] and sin 45° =
∵ cos 30° = [tex]\frac{\sqrt{3}}{2}[/tex] and sin 30° = [tex]\frac{1}{2}[/tex]
∴ cos(45 - 30)° = [tex]\frac{\sqrt{2}}{2}[/tex] × [tex]\frac{\sqrt{3}}{2}[/tex] +
∴ cos(45 - 30)° = [tex]\frac{\sqrt{6}}{4}[/tex] + [tex]\frac{\sqrt{2}}{4}[/tex] = [tex]\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]
∴ cos 15° = [tex]\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]
7.
∵ tan(α + β) = [tex]\frac{tan\alpha+tan\beta}{1-tan\alpha.tan\beta}[/tex]
∵ cos α = [tex]\frac{5}{13}[/tex] and 0° < α < 90°
- That means α is in the 1st quadrant, then all trigonometry
ratios are positive
∵ sin²α + cos²α = 1
∴ sin²α + ( [tex]\frac{5}{13}[/tex] )² = 1
∴ sin²α + [tex]\frac{25}{169}[/tex] = 1
- Subtract [tex]\frac{25}{169}[/tex] from both sides
∴ sin²α = [tex]\frac{144}{169}[/tex]
- Take √ for both sides
∴ sin α = [tex]\frac{12}{13}[/tex]
∵ tan α = sin α ÷ cos α
∴ tan α = [tex]\frac{12}{13}[/tex] ÷ [tex]\frac{5}{13}[/tex]
∴ tan α = [tex]\frac{12}{5}[/tex]
∵ sin β = [tex]\frac{3}{5}[/tex] and 0° < β < 90°
∵ sin²β + cos²β = 1
∴ ( [tex]\frac{3}{5}[/tex] )² + cos²β = 1
∴ [tex]\frac{9}{25}[/tex] + cos²β = 1
- Subtract [tex]\frac{9}{25}[/tex] from both sides
∴ cos²β = [tex]\frac{16}{25}[/tex]
- Take √ for both sides
∴ cos β = [tex]\frac{4}{5}[/tex]
∵ tan β = sin β ÷ cos β
∴ tan β = [tex]\frac{3}{5}[/tex] ÷ [tex]\frac{4}{5}[/tex]
∴ tan β = [tex]\frac{3}{4}[/tex]
- Substitute the values of tan α and tan β in the tan (α + β)
∵ tan(α + β) = [tex]\frac{tan\alpha+tan\beta}{1-tan\alpha.tan\beta}[/tex]
∴ tan(α + β) = [tex]\frac{\frac{12}{5}+\frac{3}{4}}{1-(\frac{12}{5})(\frac{3}{4})}[/tex]
∴ tan(α + β) = [tex]-\frac{63}{16}[/tex]
8.
∵ sin(α - β) = sin α cos β - cos α sin β
∴ sin(π - Ф) = sin π cos Ф - cos π sin Ф
∵ sin π = 0 and cos π = -1
∴ sin(π - Ф) = (0) × cos Ф - (-1) × sin Ф
∴ sin(π - Ф) = 0 + sin Ф
∴ sin(π - Ф) = sin Ф
9.
∵ cos 2Ф = 2 cos²Ф - 1
∵ cos Ф = [tex]\frac{1}{2}[/tex]
∴ cos 2Ф = 2 ( [tex]\frac{1}{2}[/tex] )² - 1
∴ cos 2Ф = 2 × [tex]\frac{1}{4}[/tex] - 1
∴ cos 2Ф = [tex]\frac{1}{2}[/tex] - 1
∴ cos 2Ф = [tex]-\frac{1}{2}[/tex]
10.
∵ sin 2Ф = 2 sinФ cosФ
∵ cosФ = 0.6 and 0° < Ф < 90°
- That means Ф is in the 1st quadrant and all its trigonometry
ratios are positive
∵ sin²Ф + cos²Ф = 1
∴ sin²Ф + (0.6)² = 1
∴ sin²Ф + 0.36 = 1
- Subtract 0.36 from both sides
∴ sin²Ф = 0.64
- Take √ for both sides
∴ sinФ = 0.8
- Substitute the values of sinФ and cosФ in the rule above
∴ sin 2Ф = 2(0.8)(0.6)
∴ sin 2Ф = 0.96
